 So, a subsequent question which could be asked is that, now we know how to find pressure in waveguide as sound is travelling through a waveguide, but what about velocity, what is the velocity of the wave or velocity of particles as sound is getting propagated in the media, especially in context of a waveguide, so for that now we are going to develop velocity equation and we are going to find its relationship with pressure and through that mathematics will figure out how a pressure and velocity related. So, the 1 d wave velocity equation is like this, so here velocity is a function of x and time, so del to u over del x square equals 1 over c square times second derivative of u with respect to time. So, this is my 1 dimensional wave equation for velocity, my 1 dimensional wave equation for velocity and the form of this wave equation is pretty much same as the 1 dimensional wave equation for pressure, so here also I have del to p over del x square equals 1 over c square times del to p over del t square and for the velocity equation the form is exactly the same. The derivation for this velocity equation is extremely similar to the derivation for the pressure equation which we have gone through earlier, so it will be helpful for you as a listener and as a student of this course that you go back and actually derive this wave equation using similar approach which we used for deriving the pressure wave equation. Now, just as the solution for a wave equation for pressure was or could be written as f 1 t minus x over c plus f 2 t plus x over c, we can by analogy also develop a solution for this velocity wave equation for this velocity wave equation. So, let us write that down, so by analogy u of x t I can write it as f 1 t minus x over c plus f 2 t minus x over c and so on and so forth plus f a 1 t minus x over c plus f a 2 t minus x over c or excuse me these should positive and so on and so forth or alternatively as we had shown that this could be written as a real component of a sum of complex variables. We can write a similar relation for velocity as u of x t is the real component of u plus e s t minus x over c, so again u plus could depend on s plus u minus e s t minus plus x over c and once again u plus and u minus could depend on frequency as was the case where p plus and p minus could depend on frequency. So, u plus is actually a function of s and so is u minus is a function of s as well. So, now as I did for the case of a pressure wave equation I can rewrite this as real of and I take e s t out u plus, so let me put a different type style of parent brackets here u plus e minus s x over c plus u minus e plus s x over c and I can rewrite this term in curly brackets as u of x and s, so it gets u which is a function of x and s times e to the power of s t. So, till so far what we are seeing is that the mathematics for developing relations for velocity and for pressure are extremely similar I am just replacing u with p and so on and so forth. So, now what I am going to do is I am going to connect this u with the pressure and see how pressure and velocity are related and for this I have to use the momentum equation which I developed in the last class and it is that momentum equation which connects u with pressure velocity with pressure. So, from the momentum equation we know that partial derivative of pressure with respect to x is nothing but negative rho naught partial derivative of velocity with respect to time and once again here p is a function of x and t x and t as well. So, let us label these equations. So, let us call this equation 1 equation 2 and this is equation 3. So, if I combine these equations what I get is from the right side. So, before I start combining let me just rewrite the equation for pressure as well which we developed here this equation. So, I will rewrite this and I use that equation and I combine with these equations and what I get is from the right side I differentiate the equation for pressure and what I get is real of minus p plus which again depends on s times e minus s x over c plus p minus e to the power of s x over c this entire thing multiplied by e s t times s over c. So, what I am doing here what I just did here is that I took this relation where for pressure differentiated this relation with respect to x and plugged it in into the left hand side of the momentum equation. So, that is what variation of pressure with respect to x looks like and this is equal to minus rho naught and now I am going to do the something similar to velocity. So, I am going to differentiate this equation equation 1 with respect to time partially differentiate it with respect to time. So, what I get is minus rho naught times real of u plus e minus s x over c plus u minus e s x over c this entire thing times s e s t the whole thing into rectangular brackets. So, as this entire equation is valid for all values of time and all valid values of x this equation can hold true only if the terms related to minus s x over c on left side are exactly the coefficients of minus s x over e minus s x over c are same as coefficients of e minus s x over c on the right side and so on and so forth. From this understanding we conclude by equating appropriate terms from left side to right side and so on and so forth that p plus is equal to rho naught c u plus and p minus equals rho naught c u minus, but there is a negative sign before that. So, from this I can calculate u plus in terms of p plus or p plus in terms of u plus and the same thing I can do with u negative. So, this is my equation number 4 thus I can rewrite my transmission line equations the following form. This is the complex amplitude for pressure and that is p plus which depends on frequency times e minus s x over c plus p minus e to the power of s x over c. Similarly, u of x s which is complex amplitude for velocity is p plus over I am introducing a new term z naught which I will define later times e minus s x over c minus p minus over z naught e to the power of s x over c, where z naught is same as rho naught c. This term z naught which is equal to rho naught over c is called characteristic impedance it is called characteristic impedance. So, from these equations I can now write the full form for pressure and velocity. So, p is a function of x and time equals real of p plus e minus s x over c plus p minus e s x over c times e to the power of s t and u which depends on again x and time is real of p plus e minus s x over c over z naught minus p negative over z naught e s x over c e to the power of s t. These two equations are called transmission line equations for sound ducts of constant cross section. So, from these two equations you can calculate pressure and you can also calculate velocity of sound as it travels through a wave guide which has a constant cross section. So, now what we will do is we will do a couple of examples. So, that things become clearer to us. So, let us do an example. So, I have again a tube, but unlike the last case here the tube is of a finite length and also the tube is closed at the extreme end. So, let us put a coordinate system here. I am measuring x from this location and the value of x is 0 here. The open end of the tube is where x equals minus l. I have a piston and this piston is generating pressure wave. So, let us say excuse me this circle could be drawn better. So, this circle is rotating like this and as this circle is rotating this piston is moving back and forth because of this particular linkage and let us say it generates a positive pressure wave of P plus such that at x equals l the strength of the wave is P plus bar the magnitude of it times cosine omega t plus phi. So, this is my boundary condition. This is one boundary condition. The other boundary condition is that at x equals 0 I have a very rigid wall. So, no wave cuts across this wall and the wall does not move at all. So, essentially what that means is that the velocity at x equals 0 is 0. So, given these two boundary conditions what we have to find is the pressure. So, this is boundary condition one B C 2 is u 0 t equals 0. So, these are the two boundary conditions and now we have to figure out what is the value of P x t for x less than 0. So, the length of the tube is l. So, for this entire length how is pressure changing with respect to time in x. So, let us rewrite this first boundary condition. So, P minus l t equals real of P plus bar which is a constant number times E j phi times E j omega t. So, all what I have done is I have rewritten the same expression in this format and when I take its real value I get the same thing back. So, from this we find that s equals j omega we get this that s equals j omega. So, now we plug this and put it into the equation for pressure. So, P x t is real of P plus e to the power of minus x over c plus P minus e to the power of s x over c E s t and now I know that s equals j omega. So, I get real of P plus j omega x over c plus P minus E j omega x over c E j omega t the whole thing in rectangular brackets. So, let us call this equation 1. Similarly, I can write the equation for velocity u of x t is real of E j omega t times P plus over z naught E minus j omega x over c minus P minus over z naught E j omega x over c. Now, I know that at x equals 0 at x equals 0 u of 0 t equals 0 because this is a rigid wall. So, the wall is not moving. So, whatever is the wall the velocity of the wall is going to be the same as velocity of fluid particles. So, velocity of fluid particles is also 0 at x equals 0. So, imposing this condition what I get is 0 equals real of E j omega t P plus P minus and then I can take z naught out. So, from this particular boundary condition this can be true only if P plus minus P minus is by itself 0 because it has to hold valid for all values of time. So, that gives me P plus equals P minus. So, the implication of a rigid boundary condition in a waveguide is that P plus is equal to P minus. So, let us label again some equations. So, that is equation 2 and equation 3. So, I am putting equation 3 back into the first equation. So, what I get is P x t equals real of P plus. So, P plus here is same as P minus. So, I take P plus out E minus j omega x over c plus E j omega x over c times E j omega t. Now, I know from my understanding of complex variables that the term within the curly brackets which is E minus j omega x over c and E plus j omega x over c when I add these two terms up I essentially get situation where sin terms get cancelled out and the cosine term itself remains. So, what I get is real of P plus times 2 cosine omega x over c times E j omega t. So, I take cosine omega x over c out of the rectangular brackets. So, 2 cosine omega x over c times real of P plus P plus times E j omega t. Now, if so let us call this equation 4. Now, if I put the value of x at minus l or if I try to figure out what is the value of P plus from this relation then what I get is my final solution for pressure is P x of t equals 2 cosine omega x over c times real of P plus bar E j phi times E j omega t and when I take the real value of this what I get is 2 cosine omega x over c times P plus bar times cosine of excuse me should not be j here cosine of phi plus omega t. So, this is the final expression for pressure in this tube as sound is travelling along this tube and this is the case for a standing wave. Likewise excuse me I can use this understanding and the fact that P plus is equal to P minus and I can also develop a relation for velocity using equation 2. So, I can now I have figured out what is the relation for pressure for all values of x in time and I can also figure out what is the value of u for all values of x in time. So, let us summarize these two relations. So, my final answer is P of x t equals 2 P plus bar this thing has to have its magnitude times cosine omega x over c times cosine omega t plus phi and velocity is equal to 2 P plus bar over rho naught c sin omega x over c times sin omega t plus phi. It turns out that this is also the relation for standing waves. So, what is happening here is that you are having pressure being generated at one end sound is travelling it hits the other wall it hits the wall at x equals 0 gets reflected and once things have stabilized in this tube because the steady state solution which this whole the pressure wave equation is giving us you get standing waves for pressure and also for velocity and the way these waves vary with respect to x we will draw it here. So, here my x equals 0 let us say I am going to plot in the negative direction because my open end of the tube is at x equals minus l and I will draw two envelopes. So, at x equals 0 my pressure is maximum. So, all I am going to right now what I am going to plot is only this portion of pressure and only this portion of velocity. So, at x equals 0 my pressure is maximum and let us say that maximum now that maximum could vary with because of time fluctuation. So, it could vary between these two limits then at x equals pi excuse me at x equals lambda over 2 that pressure goes down to its negative value and then after at lambda it again comes down to its positive maximum. So, it varies something like this. So, let us say here x equals lambda over 2. So, it becomes 0 and excuse me this is not a good depiction. So, let us try to do it better. So, it varies like this and the other mirror image of this envelope is something like this. So, this is called spatial envelope for pressure and this spatial envelope for pressure is depicted by the expression 2 p plus bar times cosine omega x over c. For velocity at x equals 0 the velocity is 0 at this point. So, wherever you have velocity is minimum pressure is maximum and wherever pressure is minimum velocity is maximum. So, the spatial envelope for velocity could be depicted by the relation 2 p plus bar over z naught sin omega x over c wherever I have a minimum of pressure I will have a maximum or maximum for velocity and so on and so forth. So, if I have to plot the spatial envelope for velocity the magnitude of that spatial envelope for velocity is going to be 2 p plus over z naught. So, it is going to be because z naught is more than 1. So, it is going to be lesser than the spatial envelope for pressure. So, let us say that envelopes are limited by this blue dotted line. So, it is going to be maximum here. So, it is going to vary like this and I can keep on extending it backwards as x grows on the negative side. So, this blue curve is once again spatial envelope for velocity and we see that it is the envelopes magnitude is 0 at x equals 0 because we have a rigid wall there and that essentially injects a rigid wall boundary condition which implies that velocity of particles of fluid is 0 at that specific location. With that I wanted to close this lecture, but also but before I do that I like to introduce 2, 3 more terms which will be using in later lectures. So, in context of standing waves we found that there are places or there are locations where you have a null that is. So, the first term I am introducing is null and more specifically you can also call it as a velocity null and it is a place where velocity is 0. So, these points these are all velocity nulls corresponding to velocity nulls you have locations corresponding to velocity nulls you have pressure maximum. So, pressure is maximum where you have a velocity null. Now, these points where you have velocity as maximum there the pressure is minimum. So, these points are called nodes and these are where velocity is max or pressure is minimum. In case of a tube which terminates with a rigid wall the value of minimum pressure is exactly 0 and the value of velocity is also exactly 0, but if you have a termination condition where it is not an absolute case that is the termination condition is such that the wall is not absolutely rigid then the pressure may be still at a minimum, but it may not necessarily be exactly at a 0 value and same thing holds true for velocity as well. So, the reason these nulls or minima values for velocity and pressure are exactly 0 is because a we have a rigid wall termination condition and the second reason is that there is no damping happening in the system. I also wanted to introduce three couple of more terms. So, we have talked about characteristic impedance and characteristic impedance we defined as p plus over u minus the magnitudes of these and that is essentially Z naught and this is equal to rho naught c and the value of characteristic impedance is equal to 415 Pascal second per meter for air at 20 degree centigrade and 1 atmospheres. If we have fluid sound propagation happening in water then the value for that situation is 1.48 times 10 to the power of 6 Pascal second per meter for water at 20 degree centigrade and this is fresh water we are talking about. If I go to saline water then these numbers change and so it is important to understand the exact nature of the medium which we are using as sound is travelling through it. So, that is the first definition the second one is specific acoustic impedance and that equals so this is equal to p of x and s divided by u of x and s. So, unlike characteristic impedance which is a pure number specific acoustic impedance can change with x and it will can it can also change with s which is frequency. So, this varies with x and s and this is designated as Z and Z could be a function of we already mentioned x and s. Now, Z of x and s could be written as a real part. So, designated by the letter R and that again can depend on x and s and also an imaginary part. So, I have a J times K R which again is a function of x and s is called specific acoustic resistance and it depends on the damping parameters of the system. So, Z x and s is a property of the system which includes the medium and all the devices working in the medium. That specific in acoustic impedance can vary with respect to x and s and it can be broken up into a real part which relates to damping phenomena in the system and the real part is called specific acoustic resistance and the imaginary component is called specific acoustic reactance and that relates to compliance and inertial related parameters of the system. The final definition I wanted to introduce here is called driving point impedance and this is more used in electrical domain and this is essentially same as V which is a function of s and V standing for voltage divided by current which again depends on s and this is nothing but Z of s. So, what we have covered today is a continuation of 1 d wave propagation of sound. We have understood that its solution could be represented in very general form as f 1 t minus x over c plus f 2 t minus t plus x over c and we have understood the physical significance of f 1 and f 2 and then after that we have talked about waveguides and transmission line equations as in the case of sound propagating through ducts of uniform cross section and then we have done a couple of examples and we have also understood how standing waves get created in a tube which is of a finite length and which terminates rigidly. So, with this I close today's lecture and we will continue our journey of understanding 1 d sound propagation in the next lecture. Thank you very much.